I have been thinking for an hour and i still couldn't get the answer for this qns. It involve moment of inertia.(adsbygoogle = window.adsbygoogle || []).push({});

a uniform solid disk, radius R and mass M, free to rotate about a frictionless pivot on a point on its rim.It is released from a position where the centre of mass is horizontal to its pivot, and its allowed to swing till the centre of mass is vertically below. What is the speed of the centre of mass when is at that position?

My solution

moment of inertia=Icm+MR^2

moment of inertia= 1/2 MR^2 +MR^2=3/2 MR^2

Torque=moment of inertia X angular a

Mg R= 3/2 MR^2 a

a=2g/3R

Final angular v (Wf)^2= Wi^2 +2a(angular displacement)

Wf^2= 0 +2(2g/3R)(pi/4)

Wf can be found

V of the centre of mass , R distance away from the pivot= RWf

However, the answer in the book is 2 (Rg/3)^0.5

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# Question on moment of inertia

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